Math 409/409G History of Mathematics Perfect Numbers

What’s a perfect number? A positive integer is a perfect number if it is equal to the sum of all its positive divisors except itself. For example, 6 is a perfect number since the positive divisors of 6 are 1, 2, 3, and 6 and 6  (the sum of all positive divisors of 6 except 6).

In fact, P 1  6 is the smallest perfect number. The next in line is P 2  28. (The positive divisors of 28 are 1, 2, 4, 7, 14, and 28 and 28  ) The third and fourth perfect numbers are P 3  496 and P 4  As another example, no prime number p can be a perfect number since the only divisors of p are 1 and p and p ≠ 1.

Why study perfect numbers? Ancient philosophers thought that perfect numbers had mystical and religious significance: God created the world in 6 days and rested on the seventh; it takes 28 days for the moon to circle the earth. We, of course, study perfect numbers for the sheer beauty of the mathematics.

How do you find the positive divisors of a number? Look at the prime factorization of the number. The positive divisors of n are of the form where

For example, 28  2 2 ·7. So the positive divisors of 28 are of the form d  2 a ·7 b where a  0, 1, or 2 and b  0 or 1. Since there are 3 choices for a and 2 for b, 28 has 3·2  6 positive divisors. With b  0 we get the divisors 2 0, 2 1, and 2 2. And with b  1 we get 2 0 ·7, 2 1 ·7, and 2 2 ·7. So the six positive divisors of 28 are 1, 2, 4, 7, 14, and 28.

Finding all positive divisors of a number can be an exhausting task. For example, if n  2 5 ·3 3 ·7 2 ·11, then when determining the positive divisors of n we have:6 choices for the exponent of two, 4 for the exponent of three, 3 for the exponent of seven, and 2 for the exponent of eleven. So n has 6·4·3·2  144 positive divisors!

To determine if n  2 5 ·3 3 ·7 2 ·11 is a perfect number we would have to find the 143 divisors other than n itself and then add them up. There’s got to be a better way! We need a formula for the sum of these divisors.

The sigma function If n is a positive integer, then σ(n) is defined to be the sum of all the positive divisors of n. Examples: The positive divisors of 6 are 1, 2, 3, and 6. So σ(6)   12. The positive divisors of 28 are 1, 2, 4, 7, 14, and 28. So σ(28)  56, the sum of all these positive divisors.

Since σ(n) is the sum of all the positive divisors of n, σ(n) – n is the sum of all positive divisors except n. So n is a perfect number when σ(n) – n  n. That is n is a perfect number if σ(n)  2n. Examples: 6 and 28 are perfect numbers since σ(6)  12  2·6 and σ(28)  56  2·28.

Theorem: Example:

Proof that For i  1, 2, 3, …, r, let Ex.For n  28  2 2 ·7 1, P 1  and P 2  Consider the product P 1 P 2 P 3 ···P r.

Ex.For n  28, P 1 P 2  ( )(1 + 7)  Each positive divisor of n appears exactly once in the expansion of this product, so σ(n)  P 1 P 2 P 3 ···P r. But is a geometric series. Thus

We now have our desired formula:

Example: Is n  2 5 ·3 3 ·7 2 ·11 a perfect number? Solution: So n is not a perfect number.

Can we generate perfect numbers? Consider the first four perfect numbers They are each of the form of 2 k ·p where p is a prime number.

Will any number of the form 2 k ·p where p is a prime number be a perfect number? No. Consider 44  2 2 ·11. So 44 is not a perfect number.

What condition must be placed on the prime p and the exponent k of 2?

We now see that the first four perfect numbers are of the form 2 k (2 k + 1 – 1). Is every number of this form a perfect number?

Consider n  2 8 (2 9 – 1)  2 8 ·7·73  130,816. So n is not perfect, and thus not all numbers of the form 2 k (2 k + 1 – 1) are perfect numbers. Which numbers of the form 2 k (2 k + 1 – 1) are perfect numbers?

The pattern we found for the first four perfect numbers was that they were of the form 2 k (2 k + 1 – 1) where 2 k + 1 – 1 is prime.

In the last example of a number that was not perfect, n  2 8 (2 9 – 1) where 2 9 – 1  7·73 is not prime. So will the numbers of the form 2 k (2 k + 1 – 1) where 2 k + 1 – 1 is prime generate perfect numbers? Yes, as shown in the next theorem.

Euclid’s Theorem: If 2 k + 1 – 1 is prime, then 2 k (2 k + 1 – 1) is perfect. Pf:Let p  2 k + 1 – 1 be prime and set n  2 k ·p. Then But p  2 k + 1 – 1, and thus p + 1  2 k + 1. So n is perfect since

So Euclid’s theorem will generate even perfect numbers. Actually, it can be shown that all even perfect numbers are of the form 2 k (2 k + 1 – 1) where 2 k + 1 – 1 is prime. So Euclid’s theorem will generate all even perfect numbers. But generating these even perfect numbers not as easy as it looks since for each k, you have to determine if 2 k + 1 – 1 is prime.

If we look at the first four perfect numbers for inspiration to help us determine when 2 k + 1 – 1 is prime, we suspect that k + 1 must be prime.

Although I will not prove it here, it has been shown that if 2 k + 1 – 1 is prime, then k + 1 must be prime. Putting this together with Euclid’s theorem shows that the every even perfect numbers is of the form 2 p - 1 (2 p – 1) where p is prime. But we are still not guaranteed that every number of the form 2 p - 1 (2 p – 1) where p is prime will be a perfect number.

Ex:p  11 is prime, but 2 11 – 1  23·89 is not prime. So n  2 10 (2 11 – 1) is not a perfect number. The proof of this example is left as an exercise.

This ends the lesson on Perfect Numbers